Problem: $\sum\limits_{k=0}^{{99}}{{{2(3)^{k}}}} \approx $ Choose 1 answer: Choose 1 answer: (Choice A) A $5.15 \cdot 10^{47}$ (Choice B) B $3.80 \cdot 10^{30}$ (Choice C) C $2.37 \cdot 10^{-30}$ (Choice D) D $7.76 \cdot 10^{-48}$
Solution: What is the question asking for? The question is asking for the sum of the values of $2(3)^k$ from $k = 0$ to $k = 99$ : $2(3)^0 + 2(3)^1 +... +2(3)^{99} $ The series is geometric because the formula $2(3)^k$ is an exponential function of $k$. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The number of terms $n$ is ${100}$ because there are ${100}$ numbers from $0$ to $99$. The first term $a_1$ is $2$ because $2(3)^0 = 2$. The common ratio $r$ is $3$ because it is the base of the exponent in $2(3)^k$. Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{100}}&=\dfrac{{2}(1-\left({3}\right)^{{100}})}{1-{3}} \\\\ S_{{100}}&=-1\left(1-({3})^{{100}}\right)\\\\ S_{{{100}}} &\approx 5.15\cdot10^{47} \end{aligned}$ The answer $5.15 \cdot 10^{47}$